Question

# Use the t-distribution to find a confidence interval for a mean μ given the relevant sample...

Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed.

A 90% confidence interval for μ using the sample results x¯=142.0, s=52.8, and n=50

Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.

point estimate =

margin of error =

The 90% confidence interval is   to   .

Solution :

Given that,

Point estimate = sample mean = = 142.0

sample standard deviation = s = 52.8

sample size = n = 50

Degrees of freedom = df = n - 1 = 49

At 90% confidence level the t is , = 1 - 90% = 1 - 0.90 = 0.10 / 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,49 = 1.677

Margin of error = E = t /2,df * (s / n)

= 1.677 * (52.8 / 50)

= 12.52

The 90% confidence interval estimate of the population mean is, - E < < + E

142.0 - 12.52 < < 142.0 + 12.52

129.48 < < 154.52

point estimate = 142.0

margin of error = 12.52

The 90% confidence interval is 129.48 to 154.52

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